Subject: Chemistry — Nuclear Chemistry
Question: $_{92}^{238}U$ is known to undergo radioactive decay to
form $_{82}^{206}{Pb}$ by emitting alpha and beta particles. A rock
initially contained $68 \times 10^{- 6}g$ of $_{92}^{238}U$. If the
number of alpha particles that it would emit during its radioactive
decay of $_{92}^{238}U$ to $_{82}^{206}{Pb}$ in three half-lives is
$Z \times 10^{18}$, then what is the value of $Z$?
Answer: $1.2 \times 10^{18}$
Solution: $_{92}^{238}U \rightarrow_{82}^{206}{Pb} + 8_{2}^{4}{He} + 6_{- 1}^{0}\beta$
$$0.286 \times 10^{- 6}mole$$
$$= 0.286 \times 6.02 \times 10^{17}$$
$$= 1.72 \times 10^{17}$$
After 3 Half lives
$$N = \frac{No}{(2)^{3}} = \frac{1.72 \times 10^{17}}{8}$$
$$= 0.215 \times 10^{17}$$
So, no. of molecule or uranium decayed
$$= \lbrack 1.72 - 0.215\rbrack \times 10^{17}$$
$$= 1.5 \times 10^{17 -}$$
So No. of $\alpha$ particle produced $= 8 \times 1.5 \times 10^{17}$
$$= 12 \times 10^{17}$$
$$= 1.2 \times 10^{18}$$